# I Why can you write the solution in this form?

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1. Jun 4, 2017

### cristianbahena

In a barrier potential with sections: I: V(x)=0 (x<-a), II:V(x)=V (-a<x<a) and III:V(x)=0 (a<x) you can write the solution in this form:

Ψ(x)=Ae^(ikx)+Be^(-ikx) (x<-a)
Ψ(x)=Ce^(ik'x)+De^(-ik'x) (-a<x<a)
Ψ(x)=Ee^(ikx) (a<x)
and with boundary conditions solve,
but why do you can write this solution in this form:

Ψ(x)=Ae^(ikx)+DAR^e(-ikx) (x<-a)
Ψ(x)=ATe^(ikx) (a<x)

with R reflection coefficient and T transmission coefficient

2. Jun 5, 2017

### Staff: Mentor

At the mathematical level, this is just the introduction of new variables, something you can always do. Define T=E/A, define C'=C/A, D'=D/A, then write C instead of C' and D instead of D', and finally introduce R=B/DA.
In terms of physics, it reflects that we are not interested in the overall magnitude of the wave function, but only the relative fraction of transmission and reflection. If everything is proportional to A, that parameter disappears if you take ratios.